# Help With Difficult Deductive Proof [closed]

Suppose we have a Gaussian beam with a complex envelope expressed by the following equation 1: $$\tag{1} A_G(x,y,z) = \frac{A_1}{q(z)} e^{-ik \frac{x^2 + y^2}{2q(z)}}$$ where $$q(z) = z+iz_0$$ and $i$ is equal to $\sqrt{-1}$.

Suppose we have another wave whose complex envelope can be expressed by equation 2 (where $X(.)$,$Y(.)$, and $Z(z)$ are real functions): $$\tag{2} A(x,y,z) = X( \sqrt{2} \frac{x}{W(z)})Y(\sqrt{2} \frac{y}{W(z)})e^{iZ(z)}A_G(x,y,z)$$

Lastly, we have equation 3, the paraxial Helmholtz equation: $$\tag{3} \Delta _T A(r) - 2ik \frac{ \partial A}{ \partial z} =0$$

where $\Delta _T=\partial_{xx}+\partial_{yy}$ is the transverse Laplace operator.

I need to show that substituting equation (2) into equation (3), given that equation (1) is also a solution of equation (3), will produce the following equation: $$\frac{1}{X} ( \frac{\partial ^{2}X}{\partial u^{2}} - 2u \frac{\partial X}{\partial u}) + \frac{1}{Y} ( \frac{\partial ^{2}Y}{\partial v^{2}} - 2v \frac{\partial Y}{\partial v}) + kW^{2}(z) \frac{\partial Z}{\partial z}=0$$ where $$u=\frac{\sqrt2x}{W(z)}$$ and $$v=\frac{\sqrt2y}{W(z)}$$ Can somebody give me some pointers as to how to begin here?

-

## closed as too localized by David Z♦Sep 30 '12 at 19:59

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

in equation (2) you mean that X and Y are function of u and v or they are simply multiplied? – Juan Sebastian Totero Sep 30 '12 at 14:30
They are functions. I will make that more clear. Thanks – John Roberts Sep 30 '12 at 14:35
Hi John - this is a site for conceptual questions, and we prefer to avoid questions that just ask how to do a problem or how to get started. In this case especially, you have a set of instructions that tells you clearly how to get started, so try doing that and see where you run into trouble. Once you can narrow your question down to focus on a particular physical concept, I'll be happy to reopen it. – David Z Sep 30 '12 at 20:01